How to show that $(x+y)^{a} \geq x^{a} + y^{a}$ given that $a \geq 1$ and $x,y \geq 0$?

Question

How to show that $(x+y)^{a} \geq x^{a} + y^{a}$ given that $a \geq 1$ and $x,y \geq 0$? Also, how to prove that the reverse inequality holds when $0 \leq a \leq 1$?

Answer 1

We deal with $a\ge 1$. If $a=1$ there is nothing to do. So let $a\gt 1$. Fix $y$, and let $f(x)=(x+y)^a-x^a-y^a$.

We have $f(0)=0$, and $f'(x)=a((x+y)^{a-1}-x^{a-1})$. So $f'(x)\ge 0$ if $x\ge 0$. It follows that for fixed $y$, $f(x)$ is non-decreasing.